Question: $\sum\limits_{n=1}^{\infty}\dfrac{2}{1+n^2}$ When applying the integral test, we get a limit that determines whether the series converges or diverges. What is this limit? Choose 1 answer: Choose 1 answer: (Choice A) A $\lim_{b\to\infty} \left[ \dfrac{6b}{3b+b^3}-\dfrac{3}{2} \right]$ (Choice B) B $\lim_{b\to\infty}2\ln ( 1+b^2 )$ (Choice C) C $\lim_{b\to\infty} \dfrac{2}{1+b^2}$ (Choice D) D $ \lim_{b\to\infty}\left[ 2 \tan^{-1}(b) - \dfrac{\pi}{2} \right]$
Answer: $\dfrac{2}{1+n^2}$ satisfies the conditions for the integral test. This means that $\sum\limits_{n=1}^{\infty} \dfrac{2}{1+n^2}$ converges/diverges together with $\int_1^{\infty} \dfrac{2}{1+x^2} \,dx$. $\int_1^{\infty} \dfrac{2}{1+x^2} \,dx=\lim_{b\to\infty} \left[ 2 \tan^{-1}(b) - \dfrac{\pi}{2} \right]$ In conclusion, the limit that determines whether the series converges or diverges is $\lim_{b\to\infty} \left[ 2 \tan^{-1}(b) - \dfrac{\pi}{2} \right]$.